A closed inexpansible vessel contains air saturated with water vapour at 77 degree Celcius. The total pressure of the vessel is 1007mmHg. Calculate the new pressure if the temperature is reduced to 27 degree celcius. (svp water at 77 degree celcius and 27 degree celcius respectively are 314mmHg and 27mmHg. Treat the air in the vessel as an ideal gas )
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To calculate the new pressure when the temperature is reduced, we can use the ideal gas law equation. The ideal gas law equation is:
PV = nRT
Where:
P = Pressure
V = Volume
n = Number of moles of gas
R = Ideal gas constant
T = Temperature
In this case, since the vessel is closed and considered an ideal gas, the volume remains constant. Therefore, we can rewrite the equation as:
P1/T1 = P2/T2
Where:
P1 = Initial pressure
T1 = Initial temperature
P2 = Final pressure (what we need to calculate)
T2 = Final temperature
Given information:
P1 = 1007 mmHg
T1 = 77°C = 350K
T2 = 27°C = 300K
Now we can plug in the values into the equation:
1007 mmHg / 350K = P2 / 300K
To solve for P2, we can cross-multiply:
1007 mmHg * 300K = 350K * P2
301,200 mmHg*K = 350K * P2
Now divide both sides by 350K to isolate P2:
301,200 mmHg*K / 350K = P2
The temperature units cancel out:
860.57 mmHg = P2
Therefore, the new pressure when the temperature is reduced to 27°C is approximately 860.57 mmHg.
Work each gas independently.
Pair=kT
so for change of air pressure..
P2/(1007-314)=(273+23)/(273+77)
solve for P2, the partial pressure of air at 23C
then you have the partial pressure of water vapor listed as 27mmHg at the new temp. so the Pressure must be P2+27mmhg